Snowstorm Stella impacted both our numbers and our location, but last night a smaller crew braved the cold weather and messy streets to celebrate Pi Day with pizza and Pi Cake at Ribalta.

We naturally ate a lot of round pies and even a rectangular pie to honor Hippocrates’ squaring the lune.

This year’s Pi Cake came from Empire Cakes for the third year in a row.  It was their Brooklyn Blackout cake with Chocolate frosting, a blue Pi symbol on top and blue circles with red radii around the sides.

Some pictures from last night:

And all the years’ Pi Cakes:

Last night we celebrated Rounded Pi Day by rounding at the 10,000’s digit to get 3.1416 which nicely works with the date 3/14/16.  This was great after Mega Pi Day worked out so perfectly last year.  And this all built upon previous years’ celebrations.

We ate a large quantity of pizza at Lombardi’s. and for the second year in a row we got the Pi Cake from Empire Cakes with peanut butter and chocolate flavors.  The base was inscribed with historic approximations of Pi:  25/8, 256/81, 339/108, 223/71, 377/120, 3927/1250, 355/113, 62832/20000, 22/7.

Some pictures from the fantastic night:

Previous year’s Pi Cakes:

A friend recently posted the following the problem:

There are 10 green balls, 20 red balls, and 25 blues balls in a a jar. I choose a ball at random. If I choose a green then I take out all the green balls, if i choose a red ball then i take out all the red balls, and if I choose, a blue ball I take out all the blue balls, What is the probability that I will choose a red ball on my second try?

The math works out fairly easily. It’s the probability of first drawing a green ball AND then drawing a red ball, OR the probability of drawing a blue ball AND then drawing a red ball.

$\frac{10}{10+20+25} * \frac{20}{20+25} + \frac{25}{10+20+25} * \frac{20}{10+20} = 0.3838$

But I always prefer simulations over probability so let’s break out the R code like we did for the Monty Hall Problem and calculating lottery odds.  The results are after the break.

Michael Malecki recently shared a link to a Business Insider article that discussed the Monty Hall Problem.

The problem starts with three doors, one of which has a car and two of which have a goat. You choose one door at random and then the host reveals one door (not the one you chose) that holds a goat. You can then choose to stick with your door or choose the third, remaining door.

Probability theory states that people who switch win the car two-thirds of the time and those who don’t switch only win one-third of time.

But people often still do not believe they should switch based on the probability argument alone. So let’s run some simulations.

This function randomly assigns goats and cars behind three doors, chooses a door at random, reveals a goat door, then either switches doors or does not.

monty <- function(switch=TRUE)
{
# randomly assign goats and cars
doors <- sample(x=c("Car", "Goat", "Goat"), size=3, replace=FALSE)

# randomly choose a door
doorChoice <- sample(1:3, size=1)

# get goat doors
goatDoors <- which(doors == "Goat")
# show a door with a goat
goatDoor <- goatDoors[which(goatDoors != doorChoice)][1]

if(switch)
# if we are switching choose the other remaining door
{
return(doors[-c(doorChoice, goatDoor)])
}else
# otherwise keep the current door
{
return(doors[doorChoice])
}
}


Now we simulate switching 10,000 times and not switching 10,0000 times

withSwitching <- replicate(n = 10000, expr = monty(switch = TRUE), simplify = TRUE)
withoutSwitching <- replicate(n = 10000, expr = monty(switch = FALSE), simplify = TRUE)


## [1] "Goat" "Car"  "Car"  "Goat" "Car"  "Goat"

head(withoutSwitching)

## [1] "Goat" "Car"  "Car"  "Car"  "Car"  "Car"


mean(withSwitching == "Car")

## [1] 0.6678

mean(withoutSwitching == "Car")

## [1] 0.3408


Plotting the results really shows the difference.

require(ggplot2)

## Loading required package: ggplot2

require(scales)

## Loading required package: scales

qplot(withSwitching, geom = "bar", fill = withSwitching) + scale_fill_manual("Prize",
values = c(Car = muted("blue"), Goat = "orange")) + xlab("Switch") + ggtitle("Monty Hall with Switching")


qplot(withoutSwitching, geom = "bar", fill = withoutSwitching) + scale_fill_manual("Prize",
values = c(Car = muted("blue"), Goat = "orange")) + xlab("Don't Switch") +
ggtitle("Monty Hall without Switching")


(How are these colors? I’m trying out some new combinations.)

This clearly shows that switching is the best strategy.

The New York Times has a nice simulator that lets you play with actual doors.

Some Pi Day pictures I’ve come across today.  Really love how so many people are getting into the holiday.

That was my favorite, more after the break.  I’ll continue updating as they catch my eye.

With tonight’s Mega Millions jackpot estimated to be over \$640 million there are long lines of people waiting to buy tickets.  Of course you always hear about the probability of winning which is easy enough to calculate:  Five numbers ranging from 1 through 56 are drawn (without replacement) then a sixth ball is pulled from a set of 1 through 46.  That means there are choose(56, 5) * 46 = 175,711,536 possible different combinations.  That is why people are constantly reminded of how unlikely they are to win.

But I want to see how likely it is that SOMEONE will win tonight.  So let’s break out R and ggplot!

As of this afternoon it was reported (sorry no source) that two tickets were sold for every American.  So let’s assume that each of these tickets is an independent Bernoulli trial with probability of success of 1/175,711,536.

Running 1,000 simulations we see the distribution of the number of winners in the histogram above.

So we shouldn’t be surprised if there are multiple winners tonight.

The R code:

winners <- rbinom(n=1000, size=600000000, prob=1/175000000)
qplot(winners, geom="histogram", binwidth=1, xlab="Number of Winners")

Shortly after the Giants fantastic defeat of the Patriots in Super Bowl XLVI (I was a little disappointed that Eli, Coughlin and the Vince Lombardi Trophy all got off the parade route early and the views of City Hall were obstructed by construction trailers, but Steve Weatherford was awesome as always) a friend asked me to settle a debate amongst some people in a Super Bowl pool.

He writes:

We have 10 participants in a superbowl pool.  The pool is a “pick the player who scores first” type pool.  In a hat, there are 10 Giants players.  Each participant picks 1 player out of the hat (in no particular order) until the hat is emptied.  Then 10 Patriots players go in the hat and each participant picks again.

In the end, each of the 10 participants has 1 Giants player and 1 Patriots player.  No one has any duplicate players as 10 different players from each team were selected.  Pool looks as follows:

 Participant 1 Giant A Patriot Q Participant 2 Giant B Patriot R Participant 3 Giant C Patriot S Participant 4 Giant D Patriot T Participant 5 Giant E Patriot U Participant 6 Giant F Patriot V Participant 7 Giant G Patriot W Participant 8 Giant H Patriot X Participant 9 Giant I Patriot Y Participant 10 Giant J Patriot Z

Winners = First Player to score wins half the pot.  First player to score in 2nd half wins the remaining half of the pot.

The question is, what are the odds that someone wins Both the 1st and 2nd half.  Remember, the picks were random.

Before anyone asks about the safety, one of the slots was for Special Teams/Defense.

There are two probabilistic ways of thinking about this.  Both hinge on the fact that whoever scores first in each half is both independent and not mutually exclusive.

First, let’s look at the two halves individually.  In a given half any of 20 players can score first (10 from the Giants and 10 from the Patriots) and an individual participant can win with two of those.  So a participant has a 2/20 = 1/10 chance of winning a half.  Thus that participant has a (1/10) * (1/10) = 1/100 chance of winning both halves.  Since there are 10 participants there is an overall probability of 10 * (1/100) = 1/10 of any single participant winning both halves.

The other way is to think a little more combinatorically.  There are 20 * 20 = 400 different combinations of players scoring first in each half.  A participant has two players which are each valid for each half giving them four of the possible combinations leading to a 4 / 400 = 1/100 probability that a single participant will win both halves.  Again, there are 10 participants giving an overall 10% chance of any one participant winning both halves.

Since both methods agreed I am pretty confidant in the results, but just in case I ran some simulations in R which you can find after the break.

The FBI has put out a public request for help cracking a code.  The code above was found in the pants of a murder victim over 10 years ago.  Despite some of the best code breakers in the world give it a shot, they have not been able to break the code.  I wonder if the NSA had a go at it.  Couldn’t they try brute force like in Dan Brown’s Digital Fortress?  Yes I referenced Dan Brown in the same paragraph as the NSA, deal with it.

If you think you can help send a letter to:

FBI Laboratory
Cryptanalysis and Racketeering Records Unit
2501 Investigation Parkway
Quantico, VA 22135
Attn: Ricky McCormick Case

There’s no reward but you’d be helping your country.

As mentioned earlier, yesterday was Pi Day so a bunch of statisticians and other such nerds celebrated at the new(ish) Artichoke Basille near the High Line.  We had three pies:  the signature Artichoke, the Margherita and the Anchovy, which was delicious but only some of us ate.  And of course we had our custom cake from Chrissie Cook.

The photos were taken by John.